GIF

Sum of the even terms:

$S_{even}= \sum_{m=1}^{20} \left \lfloor \frac{(2m)^2}{2} \right \rfloor= \sum_{m=1}^{20} \left \lfloor \frac{4m^2}{2} \right \rfloor =\sum_{m=1}^{20} \left \lfloor 2m^2 \right \rfloor =\sum_{m=1}^{20} 2m^2 =2\sum_{m=1}^{20} m^2 =2\left(\frac{20^3}{3}+\frac{20^2}{2}+\frac{20}{6}\right)=5740$

Sum of the odd terms:

$S_{odd}=\sum_{m=1}^{20} \left \lfloor \frac{(2m-1)^2}{2} \right \rfloor= \sum_{m=1}^{20} \left \lfloor \frac{4m^2-4m+1}{2} \right \rfloor =\sum_{m=1}^{20} \left \lfloor 2m^2 -2m +\frac{1}{2} \right \rfloor =\sum_{m=1}^{20} (2m^2-2m) =5740-2\sum_{m=1}^{20} m =$

$=5740-2\left(\frac{20^2}{2}+\frac{20}{2}\right)=5740-2\times210=5320$

Overall sum is a sum of the even and the odd:

$M=S_{even}+S_{odd}=5740+5320=\boxed{11060}$

Similar solution with @Marta Reece 's

$\begin{aligned} M & = \sum_{n=1}^{40} \left \lfloor \frac {n^2}2 \right \rfloor \\ & = \underbrace{\sum_{k=1}^{20} \left \lfloor \frac {(2k)^2}2 \right \rfloor}_{n = 2k \text{ even}} + \underbrace{\sum_{\color{#3D99F6}k=0}^{19} \left \lfloor \frac {(2k+1)^2}2 \right \rfloor}_{n = 2k+1 \text{ odd}} & \small \color{#3D99F6} \text{Note that } \left \lfloor \frac {(2{\color{#D61F06}(0)}+1)^2}2 \right \rfloor = \color{#D61F06}0 \\ & = \sum_{k=1}^{20} \left \lfloor 2k^2 \right \rfloor + \sum_{\color{#D61F06}k=1}^{19} \left \lfloor 2k^2 + 2k + \frac 12 \right \rfloor \\ & = \sum_{k=1}^{20} 2k^2 + \sum_{k=1}^{19} \left(2k^2 + 2k\right) \\ & = \frac {2(20)(21)(41)}6 + \frac {2(19)(20)(39)}6 + \frac {2(19)(20)}2 \\ & = 5740 + 4940 + 380 \\ & = \boxed{11060} \end{aligned}$